Were I to fall towards a typical black hole, the tidal forces would rip me apart well before I got to the event horizon. However, if a black hole were big enough, I could enter the event horizon before tidal forces ripped me apart. How big would this black hole have to be in terms of mass and Schwarzschild radius?

Unless you have better numbers, let's say I don't want to suffer more than 10 Gs of tidal force.

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    $\begingroup$ Er, I don't have an answer, but I just want to point out some fact just for fun if you don't mind. The black hole in the movie Interstellar does not have enough tidal force. $\endgroup$ – TBBT Jul 13 '15 at 0:45
  • $\begingroup$ Are you in a space ship or just your suit? Actually, it doesn't matter. You will experience more than 100 Gs of tidal force before you get to the event horizon. You do realize that the G force at the event horizon will accelerate small particles to 1/2 the speed of light? $\endgroup$ – LDC3 Jul 13 '15 at 0:46
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    $\begingroup$ @LDC3 A large black hole will have a smaller gradient. It would also accelerate you over a long period of time. $\endgroup$ – PyRulez Jul 13 '15 at 0:59
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    $\begingroup$ @LDC3 If you have calculations showing that the tidal forces always exceed 10 Gs, I would gladly accept it. $\endgroup$ – PyRulez Jul 13 '15 at 1:24
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    $\begingroup$ I am not sure that I would consider this question as biophysics. $\endgroup$ – Kyle Kanos Jul 13 '15 at 1:47

This problem is dealt with (in the context of classical General Relativity) nicely in Taylor & Wheeler's book "Exploring black holes: An introduction to General Relativity" (2000, Addison, Wesley, Longman).

In the section entitled "Project B: Inside the black hole" they perform a calculation for a free-falling observer, based on the Schwarzschild metric for non-rotating black holes, for the time it will take from being "uncomfortable" to reaching the singularity at the centre and the radius at which this occurs.

It turns out that this time is independent of the mass of the black hole and is equal to $$ \tau = \frac{2c}{3}\left(\frac{\Delta r}{g}\right)^{1/2},$$ where $\Delta r$ is your height and $g$ is the differential acceleration you are going to experience between head and feet. The radius at which this occurs does depend on black hole mass $M$ and is given by $$ r = \left( \frac{2GM \Delta r}{g}\right)^{1/3}$$

So if we equate the latter with the Schwarzschild radius $r_s = 2GM/c^2$, then the tidal "ripping" (!) takes place prior to reaching the event horizon if the black hole mass is less than $$ M < \frac{c^3}{2G} \left(\frac{\Delta r}{g}\right)^{1/2}$$

This appears to be precisely the result obtained by Alan Rominger using Newtonian gravity!

If we let $\Delta r=2$ m and $g = 100$ m/s$^2$, then $M<2.86\times 10^{34}$ kg (or $1.43\times10^{4} M_{\odot}$). More massive than this and (according to classical GR) you would be torn apart after falling inside the event horizon but before reaching the singularity.

  • $\begingroup$ Very interesting, but perhaps you could phrase it as a condition required for NOT being ripped apart? i.e. as If $M>\cdots$, you won't be ripped apart $\endgroup$ – innisfree Jul 13 '15 at 11:55
  • $\begingroup$ "but before reaching the singularity." I think it safe to say that once I reach the singularity even more ripping will occur. $\endgroup$ – PyRulez Jul 13 '15 at 17:34
  • $\begingroup$ @PyRulez You can't reach the singularity (intact), because the tidal forces become infinitely large (for a Schwarzschild hole and classical GR). $\endgroup$ – ProfRob Jul 13 '15 at 17:37
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    $\begingroup$ What is M with the circle and dot subscripted to it? $\endgroup$ – B T Dec 1 '15 at 23:44
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    $\begingroup$ @BT A solar mass. $\endgroup$ – ProfRob Dec 2 '15 at 7:56

Taking this as a matter of Fermi estimation, I will take the Newtonian form of gravity. No, this isn't great accuracy, but if anyone has any severe theoretical issues to raise, I will be glad to hear them. I will assume that your body extends 1 m out from its center of mass and that the extremities there will experience 10 g before your fingernails bleed and you are pronounced dead.

$$ r_s = \frac{2 GM }{c^2 } \\ \text{ tidal } = 10 g = 2 \frac{ GM}{r_c^3 } \Delta r \\ M = \frac{c^3}{2 G} \sqrt{ \frac{ \Delta r }{ 10 g } } $$

Google can calculate this. I obtain 2e34 kg, or 10,250 solar masses. That wouldn't be the largest black hole in the Milky Way. But still large enough so that finding it would be uncommon compared to the much larger stellar mass category, all of which will kill you while our telescopes can still observe it.

  • $\begingroup$ I'll add: The idea that you can survive the tidal forces "falling into" a black hole is misguided: once you pass the event horizon, you're going to the mass in the middle. At least for noncharged, nonrotating black holes, they have some distance where the tidal force outdoes whatever limits you put on it, and the infalling mass eventually gets ripped apart by tidal stresses. (I'm not sure about rotating ones; a rotating black hole often has some sort of "donut" of mass, and maybe it could be so large and spin so fast that along the axis of rotation you could fall to the very center.) $\endgroup$ – CR Drost Jul 13 '15 at 1:52
  • $\begingroup$ @ChrisDrost The idea of surviving at all is misguided. Isn't it true though that you what outlive an exterior observers (from both your own and their reference frame)? $\endgroup$ – PyRulez Jul 13 '15 at 1:54
  • $\begingroup$ @ChrisDrost Also, the idea about the ring is interesting. $\endgroup$ – PyRulez Jul 13 '15 at 1:54
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    $\begingroup$ Yes: the exterior observer will never see you rip apart due to tidal forces. But you still will. $\endgroup$ – CR Drost Jul 13 '15 at 1:59
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    $\begingroup$ Using the Schwarzschild metric appears to give exactly the same result! $\endgroup$ – ProfRob Jul 13 '15 at 11:37

Answer to question Version 1: No one knows. We can answer this question using general relativity to give a classical description but I think there is now serious doubt that GTR describes the inside of a black hole (i.e. within the event horizon) accurately and that we shall need a full quantum theory of gravity to know what happens there.

But the classical description is as follows.

You cannot fall into a black hole without dying: you will hit the singularity: one of the crucial characteristics of the event horizon is that the future of any world line beginning at any point inside the event horizon is a collision with the singularity. But the black hole might be big enough that you have a normal lifespan to live before you get there. Let's look at this further.

See my answer here, where I talk about world lines within a black hole using the really neat Kruskal–Szekeres co-ordinates. Notwithstanding their fearsome name and appearance, their crucial, neat, intuitive property is this: light cones on a KS chart look exactly like they do in flat Minkowsky spacetime.

So, looking at the KS diagram in my other answer, the time you have, depending on your initial speed and other factors, is of the order of a few $G\,M$, where $M$ is the black hole's Schwarzschild mass parameter: equal to half the Schwarzschild radius, thus equal to $G\,M/c^2$, or $G\,M/c^3$ expressed as a time. So, say we want this time to be of the order of $10^9$ seconds: a significant fraction of a human lifetime. This implies a black hole colossal mass of $10^9\,c^3/G$. If I have gotten my conversion from natural to SI units right, this comes out to be $3.9\times 10^{44}{\rm kg}$ or roughly $10^{14}$ solar masses. The Schwarzschild radius will thus be of the order of a human lifetime times a lightyear. By way of comparison, the black hole at our galaxy's center is a paltry four million solar masses. My estimate is, however, significantly below the estimate of the Universe's total energy, so it is in theory possible.

  • $\begingroup$ I clarified in the question that I only need to survive until I hit the event horizon. $\endgroup$ – PyRulez Jul 13 '15 at 2:19
  • $\begingroup$ @PyRulez In that case, Alan Rominger's answer is a good order of magnitude estimate. I am surprised by both answers: both Alan's and mine for the two questions are smaller than I thought they'd be. $\endgroup$ – Selene Routley Jul 13 '15 at 2:56

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